tag:blogger.com,1999:blog-8734744291040505450.post3507538392576008279..comments2020-02-28T19:52:10.833-08:00Comments on The Art of Logic: Pascal's triangle: a neat proofRussell Haneshttp://www.blogger.com/profile/13594411930757264086noreply@blogger.comBlogger18125tag:blogger.com,1999:blog-8734744291040505450.post-48354289453865884102014-04-24T17:51:23.590-07:002014-04-24T17:51:23.590-07:00It occurred to me, after all these years, that the...It occurred to me, after all these years, that there is an important, obvious connection to the Normal probability density function. The binomial distribution, which is modeled by Pascal's triangle, gives rise to the Normal distribution as n goes to infinity. So of course the probability of any individual outcome goes to zero; that is how the Normal distribution works. It only makes sense to speak of the probability of a specific range of outcomes in a Normal distribution.Russell Haneshttps://www.blogger.com/profile/13594411930757264086noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-31853882504498286632011-03-22T16:54:47.948-07:002011-03-22T16:54:47.948-07:00Yes, number theory would be the broad topic, I gue...Yes, number theory would be the broad topic, I guess.Russell Haneshttps://www.blogger.com/profile/13594411930757264086noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-12054020583063449342011-03-20T13:16:28.798-07:002011-03-20T13:16:28.798-07:00What I mean is: does this type of maths fall under...What I mean is: does this type of maths fall under a subject like number theory? If not, what other broad range in maths is it?<br /><br />Thanks for responding.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-85100944376606345242011-03-18T18:44:03.233-07:002011-03-18T18:44:03.233-07:00I would put this under the topic "infinite pr...I would put this under the topic "infinite products."Russell Haneshttps://www.blogger.com/profile/13594411930757264086noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-24352345213126248792011-03-18T18:33:45.367-07:002011-03-18T18:33:45.367-07:00What field of mathematics is this sort of maths cl...What field of mathematics is this sort of maths classified under?<br /><br />I am interested in this type of maths.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-50218385575812790532009-12-16T15:38:12.315-08:002009-12-16T15:38:12.315-08:00All of the N's in the previous post should be ...All of the N's in the previous post should be lower caseJohn Hernandezhttps://www.blogger.com/profile/00760755670819328720noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-51631855705316334032009-12-16T15:37:18.947-08:002009-12-16T15:37:18.947-08:00Sorry, my notation is terrible. I don't know h...Sorry, my notation is terrible. I don't know how to get an equation editor to output anything that I can paste into this comment.<br /><br />For the first, your comparing it to the sum of the two adjacent terms or:<br />200C100 / (200C99 + 200C101) which goes to 1/2<br />Doing the same with the next two terms, 200C98 + 200C102 yields the same result as you take it to infinity and so on.<br /><br />For the second L(N) is the product from i = 1 to n/2 of the values (2i-1)/(2i). L(N) is defined for even n. <br />For n= 6, L(n) = 1/2 * 3/4 * 5/6 = 15/48 <br />and 15/48 < 1/sqrt(6).<br /><br />By the way, how are you getting your equations into the post. Are you converting them to jpg's or bmp's?John Hernandezhttps://www.blogger.com/profile/00760755670819328720noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-8188353569037093652009-12-16T12:57:03.276-08:002009-12-16T12:57:03.276-08:00Hi John,
I'm sorry it took me so long to resp...Hi John,<br /><br />I'm sorry it took me so long to respond; it's a busy time at school.<br /><br />I'm not sure I understand either of your proposed methods correctly.<br /><br />For the first, you write "rather than comparing the center term to the sum of the row, compare it to the two adjacent terms. It's easily shown that this goes to 1/2." I must misunderstand what you're suggesting -- for example, 200 C 101 divided by 200 C 100 (next-to-middle / middle term) is .99 --the ratio of the middle terms to the next terms approaches 1, not 1/2.<br /><br />For the second proposal, I agree that the critical terms in the convergence is product [i=1 to inf.] of (2i-1)/2i, but I think L(n) is greater than 1/sqrt(n). For example, n=5, so 9/10 > 1/sqrt(5).Russell Haneshttps://www.blogger.com/profile/13594411930757264086noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-67295731823878084492009-12-09T11:20:18.934-08:002009-12-09T11:20:18.934-08:00Another possible solution:
Let L(n) = Product[i=1...Another possible solution:<br /><br />Let L(n) = Product[i=1 to n/2| (2i-1)/(2i)] for even n<br />The last limit in your proof is lim[n->inf | L(n)<br />It can be shown simply enough:<br />0 < L(n) <= 1/sqrt(n)<br /><br />So by the Squeeze (Sandwich) Theorem, your limit goes to zero.John Hernandezhttps://www.blogger.com/profile/00760755670819328720noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-28700304095663917382009-12-09T09:14:02.917-08:002009-12-09T09:14:02.917-08:00Actually a delta-epsilon proof on the limit works ...Actually a delta-epsilon proof on the limit works fine since for 0 < epsilon, you can find N such than n>N -> 0 < 1/(2^n) < epsilon.John Hernandezhttps://www.blogger.com/profile/00760755670819328720noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-16189519586703108352009-12-09T09:01:09.499-08:002009-12-09T09:01:09.499-08:00Consider this approach: rather than comparing the ...Consider this approach: rather than comparing the center term to the sum of the row, compare it to the two adjacent terms:<br /><br />lim n-> inf {C(n,n/2)/[C(n,n/2-1)+C(n/n/2+1)]}<br />= lim n->inf {C(n/n/2)/[2*C(n,n/2-1)}<br />It's easily shown that this goes to 1/2<br /><br />Do the same for the next two terms out:<br />lim n->inf {C(n/n/2)/[2*C(n,n/2-2)}<br />lim n-> inf {C(n,n/2)/[C(n,n/2-1)+C(n/n/2+1)]}<br /><br />This continues as you choose the next two terms. Since the middle term approaches 1/2 the adjacent terms and appoaches 1/2 the next adjacent terms or 1/4 the 4 closest terms, I'm sure with some effort, it can be shown that the middle term approaches 1/(2^n) the sum of the 2n "closest terms" so the fraction of the center term to the sum of the row can be shown to be < 1/(2^n) for any n we choose. <br /><br />Perhaps you can come up with a formalized version of this. It's not a nice, neat proof, but it can be done. Perhaps by contradiction assuming that the center term divided by the sum of the row is > 1/2^n for some arbitrary n and disproving this. I don't know if this helps.John Hernandezhttps://www.blogger.com/profile/00760755670819328720noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-2539243893255396172009-12-09T06:32:25.023-08:002009-12-09T06:32:25.023-08:00I'm still thinking about this and trying to fi...I'm still thinking about this and trying to find a solution!Russell Haneshttps://www.blogger.com/profile/13594411930757264086noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-83326689837253955092009-12-08T18:40:20.239-08:002009-12-08T18:40:20.239-08:00Never mind. It's true. The correct code is:
...Never mind. It's true. The correct code is:<br /><br />product=1;<br /> for(i=1; i< n; i++){<br /> <br /> product = product * (4 * i*i - 1) / (4 * i * i);<br /> <br /> }AnotherAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-12529364471501678732009-12-08T18:31:13.873-08:002009-12-08T18:31:13.873-08:00With regard to:
For n = 1 to inf, The product of ...With regard to:<br /><br />For n = 1 to inf, The product of (4n^2-1)/(4n^2) = 2/pi<br /><br />where (4n^2-1)/(4n^2)<1 for all natural numbers, n<br /><br /><br />I just wrote a little program to check this, and it seemed to be false. Here's a code fragment:<br /><br /> product = 1;<br /> for(i=0; i< k; i++){<br /> <br /> product = product * (4 * n*n - 1) / (4 * n * n);<br /> <br /> }<br /><br /><br />For large values of k, the product approaches zero.AnotherAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-7241542343636909452009-12-03T11:42:19.275-08:002009-12-03T11:42:19.275-08:00I read your proof a while back and your statement ...I read your proof a while back and your statement "seemed" right to me too. I was hesitant to accept it at face value, but I couldn't prove it one way or another. Then today I saw the mentioned product on another site and remembered this post.<br /><br />I still "believe" your final conclusion is true but the proof isn't htere just yet.<br /><br />Pascal's Traingle intrigues me and I'm interested if you had a use in mind for your conclusion.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-42298719588915760122009-12-03T11:24:24.595-08:002009-12-03T11:24:24.595-08:00Hmm, neat. You're right.Hmm, neat. You're right.Russell Haneshttps://www.blogger.com/profile/13594411930757264086noreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-29462413953735814182009-12-03T10:52:43.383-08:002009-12-03T10:52:43.383-08:00For n = 1 to inf, The product of (4n^2-1)/(4n^2) =...For n = 1 to inf, The product of (4n^2-1)/(4n^2) = 2/pi<br /><br />where (4n^2-1)/(4n^2)<1 for all natural numbers, nAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8734744291040505450.post-81126484154862149962009-12-03T08:37:39.078-08:002009-12-03T08:37:39.078-08:00"An infinite multiplication of fractions less..."An infinite multiplication of fractions less than or equal to 1 must converge to 0."<br /><br />This is a false statementAnonymousnoreply@blogger.com